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\begin{document}

\pagestyle{fancy}
\fancyhead{}
\lhead{NAME Jiatu Yan}
\chead{Numerical PDE/ODE homework \#5}
\rhead{Date 2021.6. 12}


\section*{I. Exercise 10.8}

From $\left( 10.16 \right) $ we have
\begin{equation*}
	\begin{split}
		&-rU_{i - 1}^{n+1}+\left( 1+2r \right)U_{i}^{n+1}-rU_{i+1}^{n}=
		rU_{i-1}^{n}+\left( 1-2r \right)U_{i}^{n}+ rU_{i+1}^{n}\qquad i = 2, \ldots, m - 1\\
		&-rg_0\left( t_{n+1} \right)+\left( 1+2r \right)U_{i}^{n+1}-rU_{i+1}^{n+1}=
		rg_0\left( t_n \right)+\left( 1-2r \right)U_{i}^{n+1}+rU_{i+1}^{n} \qquad i = 1\\  
		&-rU_{i - 1}^{n+1} +\left( 1+2r \right)U_{i}^{n+1}-rg_1\left( t_{n+1} \right) =
		rU_{i -1}^{n}+\left( 1-2r \right)U_{i}^{n+1}+rg_{1}\left( t_{n} \right)  \qquad i = 1\\  
	\end{split}
\end{equation*}
Move all the $g_i$ element to the right-hand side of each equation and we can rewrite the equations into matrix form
 \begin{equation*}
	 \begin{split}
	&\left[
		\begin{array}{c c c c c}
			1+2r& -r&\ldots&0&0\\
			-r&1+2r&\ldots&0&0\\
			\vdots&\vdots&\ddots&\vdots&\vdots\\
			0&0&\ldots&1+2r&-r\\
			0&0&\ldots&-r&1+2r\\
		\end{array}
	\right]
	U^{n+1}=
	\left[
		\begin{array}{c c c c c}
			1-2r& r&\ldots&0&0\\
			r&1-2r&\ldots&0&0\\
			\vdots&\vdots&\ddots&\vdots&\vdots\\
			0&0&\ldots&1-2r&r\\
			0&0&\ldots&r&1-2r\\
		\end{array}
	\right]
	U^{n} + 
	\left[
		\begin{array}{c}
			rg_{0}\left( t_n \right)+rg_{0}\left( t_{n+1} \right) \\
			0\\
			\vdots\\
			0\\
			rg_{1}\left( t_n \right)+rg_{1}\left( t_{n+1} \right) \\
		\end{array}
	\right]\\
\end{split}
\end{equation*}
Seperate I out from the two matrix and write $-r$ as $\frac{-k\nu}{2h^2}$, we can get the equation in (10.17), which is
\[
	\left( I-\frac{k}{2}A \right)U^{n+1}=\left( I+\frac{k}{2}A \right)U^{n}+b^{n}  
.\] 


\section*{II. Exercise 10.11}

Like Example 10.10, we have
\begin{equation*}
	\begin{split}
		\tau\left( x, t \right)=&
		\frac{u\left( x,t+k  \right)-u\left( x, t \right)  }{k}-\frac{\nu}{2h^2}
		\left[u\left( x-h,t \right)-2u\left( x, t \right)+u\left( x+h,t \right)
		+u\left( x-h,t+k \right)-2u\left( x,t+k \right)+u\left( x+h, t+k \right)   \right]\\
		=&\left( u_t+\frac{k}{2}u_{tt}+\frac{k^2}{6}u_{t t t}+\frac{k^{3}}{4!}u_{t t t t}+\ldots \right)
		 -\frac{\nu}{2h^2}\left( h^2u_{xx}+\frac{h^{4}}{4!}u_{x x x x}+\frac{h^{6}}{6!}u_{x x x x x x}+\ldots \right)\\
		 &-\frac{\nu}{2h^2}\left( h^2u_{x x}\left( x, t+k \right)+\frac{h^{4}}{4!}u_{x x x x}\left( x, t+k \right)   
		 +\frac{h^{6}}{6!}u_{x x x x x x}\left( x, t+k \right)+\ldots \right)\\
		=&\left(\nu u_{x x}+\frac{k\nu^2}{2}u_{x x x x}+\frac{k^2}{6}u_{t t t}+
			\frac{k^{3}}{4!}u_{t t t t} +\ldots\right)
		 -\frac{\nu}{2h^2}\left( h^2u_{xx}+\frac{h^{4}}{4!}u_{x x x x}+\frac{h^{6}}{6!}u_{x x x x x x}+\ldots \right)\\
		 &-\frac{\nu}{2h^2}\left( h^2u_{x x}+h^2ku_{x x t}+\frac{h^{4}}{4!}u_{x x x x}+
		 \frac{h^{4}k}{4!}u_{x x x x t}+\ldots \right)\\
		=&\frac{k^{2}}{6}u_{t t t}-\frac{\nu h^2}{4!}u_{x x x x} + \mathrm{O}\left( k^{3}+h^{4} + h^2k \right),\\ 
	\end{split}
\end{equation*}
The second line is achieved by Taylor expansion and the third one is achieved by (10.3).
Thus we have
\[
	\tau\left( x, t \right)=\mathrm{O}\left( k^2+h^{2} \right)  
.\] 

\section*{III. Exercise 10. 25 Prove the necessity part of Theorem 10.22.}

For a given consistent and convergent linear MOL, if it is not Lax-Richtmyer stable, in orther words, 
$\forall T$ and $\forall C_T>0$ there exists $k>0$, $n\in\mathbb{N}^{+}$ and $kn\le T$ s.t. 
\[
	\parallel B\left( k \right)^{n}\parallel> C_T 
.\] 
We can find two sequences $\{k_i\}$ and $\{n_i\}$ such that
\[\parallel B\left( k_i \right)^{n_i}\parallel>i .\] 
For each fixed $k_i$, we choose $n_i$ properly to satisfy $\forall 0\le Mk_i\le T$, 
$\parallel B(k_i)^{M}\parallel\le \parallel B\left( k_i \right)^{n_i}\parallel $. 
Since $\forall i=1,2,\ldots$, $0\le k_in_i\le T$ is bounded, we can get $t\in\left[0, T\right]$ and the subsequence of
$\{k_{i_j}\}$ and its cooresponding $\{n_{i_j}\}$such that  
\[
	\lim_{j\to \infty}n_{i_j}k_{i_j}=t
.\] 
For the sake of convenience, we denote the two subsequences by $\{k_i\}$ and $\{n_i\}$.
Due to the continuity of the matrix $B\left( k \right) $ about k (it is linear) and $k\in\left[0, T\right]$ is bounded, we can easily have
all the elements in  $B\left( k \right) $ are bounded.
Thus for fixed n, $\parallel B\left( k \right)^{n} \parallel$ is bounded.
So in order to let $\parallel B\left( k_i \right)^{n_i}\parallel $ tend to infinity, we must have 
 \[
\lim_{i\to \infty}n_i=\infty
,\] 
which implies 
 \[
\lim_{i\to \infty}k_i=0
.\] 
For each $k_i$, we choose $E^{0}_i$ satisfying $\parallel E^{0}_i\parallel=1$ and
\[
	\parallel B\left( k_i \right)^{n_i}E^{0}_i\parallel>
	\frac{1}{2}\parallel B\left( k_i \right)^{n_i}\parallel
\]
properly.
From the equation about global error, set $k=k_i$ we have
\begin{equation}
	\label{eq0}
	\begin{split}
		\parallel E^{n_i}\parallel&=\parallel B\left(k_i  \right)^{n_i}E^{0}_i-
		k_i\sum_{n=1}^{n_i}B\left( k_i \right)^{n_i-n}\tau^{n-1}\parallel\\
					  &\ge \mid \parallel B\left( k_i \right)^{n_i}E^{0}\parallel-\parallel k_i\sum_{n=1}^{n_i}B\left( k_i \right)^{n_i-n}\tau^{n-1}\parallel \mid    
	\end{split}
\end{equation} 
Since the MOL is consistent, we can find $K$ and $H$ such that when $k<K$ and  $h<H$ we have $\tau<\frac{1}{4T}$.
And we can find $M>0$ such that for all $i>M$, $k_i<K$.
Then for the second element in the equation (\ref{eq0}), we have
\begin{equation*}
	\begin{split}
		\parallel k_i\sum_{n=1}^{n=n_i}B\left( k_i \right)^{n_i-n}\tau^{n-1}\parallel
		&\le k_in_i\frac{1}{4T} \max_{n=1,2,\ldots,n_i}\parallel B\left( k_i \right)^{n_i}\parallel\\
		&\le \frac{1}{4}\parallel B\left( k_i \right)^{n_i}\parallel.\\ 
	\end{split}
\end{equation*}
Thus for $i>\max\{N_1, N_2\}$ we have
\[
	\parallel E^{n_i}\parallel \ge \frac{1}{4}\parallel B\left( k_i \right)^{n_i}\parallel>\frac{i}{4} 
.\] 
Since $\{E^{0}_i\}$ is a bounded set of infinite points within $\mathbb{R}^{m}$, we can choose a subset of them which 
have $\lim_{j\to 0} E^{0}_{i_j}=E^{0}$.
Set $E^{0}$ the initial error and let $j$ tend to infinity, we have
\[
\lim_{j\to \infty}\parallel E^{n_{i_j}}\parallel=\lim_{k_i\to 0\atop n_ik_i\to t}\parallel E^{n_{i_j}}\parallel \to \infty.
,\] 
which contradicts with the condition that the MOL is convergent.
Thus we proved that the convergent and consistent linear MOL is Lax-Richtmyer stable. 
\section*{IV. Exercise 10.28 Show that the modulus of the Crank-Nicolson method's amplification factor is never greater that 1 for any choice of $k,h>0$.}
Like Example 10.27, we set 
\[U_{j}^{n}=\left[g\left( \xi \right) \right]^{n}e^{ix_j\xi}.\]
And we expect that
\[
	U_{j}^{n+1}=g\left( \xi \right)U_{j}^{n} 
.\] 
Then from (10.16) we have 
\[
	\left( -re^{-ih\xi}+\left( 1+2r \right)-re^{ih\xi}  \right)g\left( \xi \right)U_{j}^{n}
	=\left( re^{-ih\xi}+\left( 1-2r \right)+re^{ih\xi}  \right)U_{j}^{n}   
.\] 
By Euler's Formula we have
\[
	g\left( \xi \right)=\frac{1-r\left( 2-2\cos\left( h\xi \right)  \right) }{1+r\left( 2-2\cos\left( h\xi \right)  \right) }
	=1-\frac{8r\sin^2\left( \frac{\xi h}{2} \right) }{1+4r\sin^2\left( \frac{\xi h}{2} \right) }\le 1
.\]
It is obvious that $g\left( \xi \right)=1$ if $h\xi=2n\pi$, $n\in\mathbb{Z}$. Then for $\xi\neq \frac{2n\pi}{h}$,
i.e. $\sin\left( \frac{\xi h}{2} \right)\neq 0 $, we have
\begin{equation*}
	\begin{split}
		0\le \frac{8r\sin^2\left( \frac{\xi h}{2} \right) }{1+4r\sin^2\left( \frac{\xi h}{2} \right) }
		=\frac{2}{\frac{1}{4r\sin^2\left( \frac{\xi h}{2} \right) }+1}\le \frac{2}{\frac{1}{4r}+1}\le 2.
	\end{split}
\end{equation*}
The last inequality is achieved by $r>0$.
Thus we have  $ \mid g\left( \xi \right) \mid \le 1 $, which means the modulus of its amplification factor is 
never greater than 1 for any choice of $k,h>0$.

\section*{V. Exercise 11.26 Show that the Beam-Warming method is second-order accurate both in time and in space.}
For the sake of convenience, $u\left( x, t \right) $ is denoted as $u$.
For the case of $a\ge 0$, we have
\begin{equation*}
	\begin{split}
		\tau\left( x,t \right)=& \frac{u\left( x,t+k \right) - u\left( x, t \right)  }{k} 
		+\frac{a}{2h}\left[3u\left( x, t \right)-4u\left( x-h, t \right)+u\left( x-2h, t \right)     \right]
		-\frac{a^2k}{2h^2}\left[u\left( x, t \right)-2u\left( x-k, t \right)+u\left( x-2h, t \right)    \right]\\
		=&\left[u_t+\frac{k}{2}u_{t t} + \mathrm{O}\left( k^2 \right) \right]
				      +\frac{a}{2h}\left[3u-4u+4hu_{x}-\frac{4h^2}{2}u_{x x}
				      +\mathrm{O}\left( h^{3} \right)
			      +u-2hu_{x}+\frac{4h^2}{2}u_{x x}+\mathrm{O}\left( h^{3} \right) \right]\\
		 &-\frac{a^2k}{2h^2}\left[u-2u+2hu_{x}-\frac{2h^2}{2}u_{x x}+\frac{2h^{3}}{3!}u_{x x x}+\mathrm{O}\left( h^{4} \right)
		 +u-2hu_{x}+\frac{4h^2}{2}h_{x x}-\frac{8h^{3}}{3!}h_{x x x}+\mathrm{O}\left( h^{4} \right) \right]\\
		=&u_t+au_x+\frac{k}{2}u_{t t}-\frac{a^2k}{2}u_{x x}+\mathrm{O}\left( k^2+hk+h^2 \right)\\
		=&\mathrm{O}\left( k^2+kh+h^2 \right). 
	\end{split}
\end{equation*}
The second equation is achieved by Taylor expansion on $\left( x, t \right) $, the third one only sorted the elements and the forth one is achieved by the advection equation. 
Similarly, for the case of $a<0$, we have
 \begin{equation*}
	\begin{split}
		\tau\left( x, t \right)=&\frac{u\left( x,t+k \right)-u\left( x, t \right)  }{k}
		+\frac{a}{2h}\left[-3u\left( x, t \right)+4u\left( x+h, t \right)-u\left( x+2h, t \right)   \right]
		-\frac{a^2k}{2h^2}\left[u\left( x, t \right)-2u\left( x+h, t \right)+u\left(x+2h, t  \right)   \right]\\
		=&\left[u_t+\frac{k}{2}u_{t t}+\mathrm{O}\left( k^2 \right)\right]
		+\frac{a}{2h}\left[-3u+4u+4hu_x+\frac{4h^2}{2}u_{x x}+\mathrm{O}\left( h^{3} \right)
		-u-2hu_x-\frac{4h^2}{2}u_{x x}+\mathrm{O}\left( h^{3} \right) \right]\\
		 &+\frac{a^2k}{2h^2}\left[u-2u-2hu_x-\frac{2h^2}{2}u_{x x}-\frac{2h^{3}}{3!}u_{x x x}+\mathrm{O}\left( h^{4} \right)
		 +u+2hu_x+\frac{4h^2}{2}u_{x x}+\frac{8h^{3}}{3!}u_{x x x}+\mathrm{O}\left( h^{4} \right) \right]\\
		=&u_t+au_{x}+\frac{k}{2}u_{t t}+\frac{a^2k}{2}u_{x x}+\mathrm{O}\left( k^2+hk+h^2 \right)\\
		=&\mathrm{O}\left( k^2+hk+h^2 \right).\\ 
	\end{split}
\end{equation*}
Thus we have proved that the Beam-Warming method is second-order accurate both in time and in space.

\section*{VI. Exercise 11.27}
For the case of $a\ge 0$, rewrite equation (11.28) and we have
 \[
	 U_j^{n+1}=U_j^{n}-\frac{\mu}{2}\left( U_{j}^{n}-U_{j-2}^{n} \right)-
	 \left( \frac{2\mu-\mu^2}{2} \right) \left(U_j^{n}-2U_{j-1}^{n}+U_{j-2}^{n}  \right)  
.\] 
This formula can be viewed as using forward Euler's method on the IVP system
\[
	U'\left( t \right)=B\left( t \right),  
\]
where 
\[
B=-\frac{a}{2h}\left[
	\begin{array}{c c c c c c}
		1&&&&-1&0\\
		0&1&&&&-1\\
		-1&0&1&&&\\
		  &&\ddots&\ddots&\ddots&\\
		  &&&0&1&\\
		  &&&-1&0&1\\
	\end{array}
\right]
-\frac{2ah-a^2k}{2h^2}
\left[
	\begin{array}{c c c c c c}
		1&&&&1&-2\\
		-2&1&&&&1\\
		1&-2&1&&&\\
		 &&\ddots&\ddots&\ddots&\\
		 &&&-2&1&\\
		 &&&1&-2&1\\
	\end{array}
\right]
.\]
Then for vector $w_p=e^{2\pi ipjh}$, $p=1, 2, \ldots, m+1$ and $j=3, \ldots, m+1$, we have
\begin{equation*}
	\begin{split}
		\left(Bw_p\right)_j&=-\frac{a}{2h}\left( e^{2\pi ipjh}-e^{2\pi ip\left( j-2 \right)h } \right)
		-\frac{2ah-a^2k}{2h^2}\left( e^{2\pi ipjh} - 2e^{2\pi ip\left( j-1 \right)h }+e^{2\pi ip\left( j-2 \right)h } \right)\\
				   &=\left[-\frac{a}{2h}\left( 1-e^{-4\pi iph} \right)
				   -\frac{2ah-a^2k}{2h^2}\left( 1-2e^{-2\pi iph}+e^{-4\pi iph} \right)  \right]e^{2\pi ipjh}\\
				   &=\lambda_p \left( w_p \right)_j.\\ 
	\end{split} 
\end{equation*}
Similarly for $j=1$ and  $j=2$.
Thus we proved that $w_p$'s are the  eigenvectors of $B$, coordinating with the eigenvalues $\lambda_p$'s respectively.
In order to let the Beam-Warming method be stable, we need to let $k\lambda_p$ all lie in the RAS of forward Euler's method.
We assume $\alpha = 2\pi ph$. $\mu$ must satisfy
\begin{equation*}
	\begin{split}
		\mid 1+k\lambda_p \mid =&  
		\mid 1-\frac{\mu}{2}\left( 1-e^{-4\pi iph} \right)-
		\left( \mu-\frac{\mu^2}{2} \right)\left( 1-2e^{-2\pi iph}+e^{-4\pi iph} \right) \mid.\\
		=&\left[\left( -\mu+\mu^2 \right)  \cos^2\left( \alpha \right)
		+\left( 2\mu-\mu^2 \right)\cos\left( \alpha\right)-\mu+1  \right]
		+\sin\left( \alpha \right) \left[\left( \mu-\mu^2 \right)\cos\left( \alpha \right)+\left( -2\mu+\mu^2 \right)   \right]i\\
	\end{split}
\end{equation*}
Now we let $x=\cos\left( \alpha \right)\in\left[-1, 1\right] $ and 
$g\left( x,\mu \right)=\left( \mu^2-\mu \right)x+\left( 2\mu-\mu^2 \right)   $, by $\cos^2+\sin^2=1$ we want to get

\begin{equation*}
	\begin{split}
		\mid 1+k\lambda_p \mid =&\left[xg\left( x, \mu \right) -\left( \mu-1 \right)  \right]^2
		+\left( 1-x^2 \right)g^2\left( x, \mu \right)\\  
		=&\mu\left( \mu-2 \right)\left( \mu-1 \right)^2x^2-2\mu\left( \mu-2 \right)\left( \mu-1 \right)^2x
		+\mu^2\left(\mu-2  \right)^2+\left( \mu-1 \right)^2  \\ 
		=&f\left(x, \mu  \right)\le1.\\ 
	\end{split}
\end{equation*}

If $\mu=0,1,2$, $f\left(x, \mu  \right) $ is a linear function about x.
If $\mu\neq 0,1,2$, its axis of symmetry is $x=1$.
$f\left( x, \mu \right) $ is nonnegative, so we only need to discuss its maximal value.
Since the range of x is $\left[-1, 1\right]$, and we can easily have 
\[
	\max_{x\in\left[-1, 1\right]}f\left( x, \mu \right)=\max\left\{f\left(-1, \mu  \right), f\left( 1, \mu \right)\right\}\\  
.\] 

When $\mu\left( \mu-2 \right) > 0 $, i.e. $\mu< 0$ or $\mu> 2$, we have
\[
	\max f\left(x, \mu  \right)=f\left( -1,\mu \right)=3\mu\left( \mu-2 \right)\left( \mu-1 \right)^2+\mu^2\left( \mu-2 \right)^2+\left( \mu-1 \right)^2      
.\] 
Because each element in the equation above are nonnegative and $\left( \mu-1 \right)\ge 1 $, this situation is wrong.

When $\mu\left( \mu-2 \right)\le 0 $, we have
\[
	\max f\left( x,\mu \right)=f\left( 1, \mu \right)=-\mu\left( \mu-2 \right)\left(\mu-1  \right)^2+\mu^2\left( \mu-2 \right)^2+\left( \mu-1 \right)^2      
.\]
From 
$
1-f\left( 1, \mu \right)=-\mu\left(\mu-2  \right)\left( \mu-1 \right)^2    
$ and $1-f\left( 1, 3 \right)=-8<0 $, we can easily have
\begin{equation*}
	\begin{split}
		\left\{
			\begin{array}{l}
				1-f\left( 1, \mu \right)<0 \qquad \mu>2 \quad\mathrm{or}\quad \mu<0\\
				1-f\left( 1, \mu \right)\ge 0\qquad 0\le \mu\le 2 
			\end{array}
			\right.
	\end{split}
\end{equation*} 
Since f is nonnegative, we have when $a\ge 0$, the Beam-Warming method is stable for $\mu\in\left[0, 2\right]$.

Similarly, for the case of $a< 0$, rewrite equation (11.29) and we have
 \[
	 U_j^{n+1}=U_j^{n}-\frac{\mu}{2}\left( U_{j+1}^{n}-U_{j}^{n} \right)+
	 \left( \frac{2\mu+\mu^2}{2} \right) \left(U_j^{n}-2U_{j-1}^{n}+U_{j-2}^{n}  \right)  
.\] 
This formula can be viewed as using forward Euler's method on the IVP system
\[
	U'\left( t \right)=B\left( t \right),  
\]
where 
\[
B=-\frac{a}{2h}\left[
	\begin{array}{c c c c c c}
		-1&0&1&&&\\
		&-1&0&1&&\\
		&&-1&&&\\
		  &&&\ddots&\ddots&\ddots\\
		  1&&&&-1&0\\
		  0&1&&&&-1\\
	\end{array}
\right]
+\frac{2ah+a^2k}{2h^2}
\left[
	\begin{array}{c c c c c c}
		1&-2&1&&&\\
		&1&-2&1&&\\
		&&1&&&\\
		 &&&\ddots&\ddots&\ddots\\
		1 &&&&1&-2\\
		 -2&1&&&&1\\
	\end{array}
\right]
.\]
Then for vector $w_p=e^{2\pi ipjh}$, $p=1, 2, \ldots, m+1$ and $j=3, \ldots, m+1$, we have
\begin{equation*}
	\begin{split}
		\left(Bw_p\right)_j&=-\frac{a}{2h}\left( e^{2\pi ip\left( j+2 \right) h}-e^{2\pi ipjh } \right)
		+\frac{2ah+a^2k}{2h^2}\left( e^{2\pi ipjh} - 2e^{2\pi ip\left( j+1 \right)h }+e^{2\pi ip\left( j+2 \right)h } \right)\\
				   &=\left[-\frac{a}{2h}\left( e^{4\pi iph}-1 \right)
				   +\frac{2ah+a^2k}{2h^2}\left( 1-2e^{2\pi iph}+e^{4\pi iph} \right)  \right]e^{2\pi ipjh}\\
				   &=\lambda_p \left( w_p \right)_j.\\ 
	\end{split} 
\end{equation*}
Similarly for $j=1$ and  $j=2$.
Thus we proved that $w_p$'s are the  eigenvectors of $B$, coordinating with the eigenvalues $\lambda_p$'s respectively.
In order to let the Beam-Warming method be stable, we need to let $k\lambda_p$ all lie in the RAS of forward Euler's method.
We assume $\alpha = 2\pi ph$. $\mu$ must satisfy
\begin{equation*}
	\begin{split}
		\mid 1+k\lambda_p \mid =&  
		\mid \frac{\mu}{2}\left( 1-e^{4\pi iph}-1 \right)+
		\left( \mu+\frac{\mu^2}{2} \right)\left( 1-2e^{2\pi iph}+e^{4\pi iph} \right) \mid.\\
		=&\left[\left( \mu+\mu^2 \right)  \cos^2\left( \alpha \right)
		+\left( -2\mu-\mu^2 \right)\cos\left( \alpha\right)+\mu+1  \right]
		+\sin\left( \alpha \right) \left[\left( \mu+\mu^2 \right)\cos\left( \alpha \right)+\left( -2\mu-\mu^2 \right)   \right]i\\
	\end{split}
\end{equation*}
Now we let $x=\cos\left( \alpha \right)\in\left[-1, 1\right] $ and 
$g\left( x,\mu \right)=\left( \mu^2+\mu \right)x-\left( 2\mu+\mu^2 \right)   $, by $\cos^2+\sin^2=1$ we want to get
\begin{equation*}
	\begin{split}
		\mid 1+k\lambda_p \mid =&\left[xg\left( x, \mu \right) +\left( \mu+1 \right)  \right]^2
		+\left( 1-x^2 \right)g^2\left( x, \mu \right)\\  
		=&\mu\left( \mu+1 \right)^2 \left( \mu+2 \right)x^2-2\mu\left(\mu+1  \right)^2\left( \mu+2 \right)  x
		+\mu^2\left(\mu+2  \right)^2+\left( \mu+1 \right)^2  \\ 
		=&f\left(x, \mu  \right)\le1.\\ 
	\end{split}
\end{equation*}
Let $\tilde{\mu} = -\mu$, we have
\[
	f\left(x, \mu  \right)= f\left( x, -\tilde{\mu} \right)=\tilde{\mu}\left( \tilde{\mu}-1 \right)^2\left( \tilde{\mu}-2 \right)x^2
	-2\tilde{\mu}\left( \tilde{\mu}-1 \right)^2\left( \tilde{\mu}-2 \right)x
	+\tilde{\mu}^2\left( \tilde{\mu}-2 \right)^2
	+\left( \tilde{\mu}-1 \right)^2 \le 1
.\]
The situation is the same as that when $a>0$, which we have discussed. We have $\tilde{\mu}\in\left[0, 2\right]$.
Thus  $\mu\in\left[-2, 0\right]$.

So we have proved that the Beam-Warming methods (11.28) and (11.29) are stable for  $\mu\in\left[0, 2\right]$ and 
$\mu\in\left[-2, 0\right]$, respectively.

The plots for  $\mu=0.8$, $1.6$, $2$ and $2.4$ are shown in Figure \ref{fig1}.

\begin{figure}[ht]
	\centering
	\caption{The figure for $z_p$ and the boundary of RAS for different  $\mu$.}
	\label{fig1}
	\begin{minipage}[t]{0.45\linewidth}
                \centering
		\includegraphics[width=8cm, height = 8cm]{Exercise11_27_1.eps}
		\caption*{$\mu=0.8$}
		\end{minipage}
	\begin{minipage}[t]{0.45\linewidth}
                \centering
		\includegraphics[width=8cm, height=8cm]{Exercise11_27_2.eps}
		\caption*{$\mu=1.6$}
	\end{minipage}
	
	\begin{minipage}[t]{0.45\linewidth}
                \centering
		\includegraphics[width=8cm, height = 8cm]{Exercise11_27_3.eps}
		\caption*{$\mu=2.0$}
		\end{minipage}
	\begin{minipage}[t]{0.45\linewidth}
                \centering
		\includegraphics[width=8cm, height=8cm]{Exercise11_27_4.eps}
		\caption*{$\mu=2.4$}
	\end{minipage}
\end{figure}

\section*{VII. Exercise 11.35 Reproduce all results in Example 11.34.}
The results are listed in Figure \ref{fig2} and Figure \ref{fig3}

\begin{figure}[ht]
        \centering
	\caption{Exercise 11.35: The final results with $k=0.8h$}
        \label{fig2} 
        \begin{minipage}[t]{0.45\linewidth}
                \centering
		\caption*{initial condition}
                \includegraphics[width=6cm, height = 6cm]{Exercise11_35_1.eps}
                \end{minipage}
        \begin{minipage}[t]{0.45\linewidth}
                \centering
		\caption*{leapfrog}
                \includegraphics[width=6cm, height=6cm]{Exercise11_35_2.eps}
        \end{minipage}

        \begin{minipage}[t]{0.45\linewidth}
                \centering
		\caption*{Lax-Friedrichs}
                \includegraphics[width=6cm, height = 6cm]{Exercise11_35_3.eps}
                \end{minipage}
        \begin{minipage}[t]{0.45\linewidth}
                \centering
		\caption*{Lax-Wendroff}
                \includegraphics[width=6cm, height=6cm]{Exercise11_35_4.eps}
        \end{minipage}

	 \begin{minipage}[t]{0.45\linewidth}
                \centering
                \caption*{upwind}
                \includegraphics[width=6cm, height = 6cm]{Exercise11_35_5.eps}
                \end{minipage}
        \begin{minipage}[t]{0.45\linewidth}
                \centering
                \caption*{Beam-Warming}
                \includegraphics[width=6cm, height=6cm]{Exercise11_35_6.eps}
        \end{minipage}
\end{figure}

\begin{figure}[ht]
        \centering
        \caption{Exercise 11.35: The final results with  k=h }
        \label{fig3} 
        \begin{minipage}[t]{0.45\linewidth}
                \centering
                \caption*{Lax-Wendroff}
                \includegraphics[width=8cm, height = 8cm]{Exercise11_35_7.eps}
                \end{minipage}
        \begin{minipage}[t]{0.45\linewidth}
                \centering
                \caption*{leapfrog}
                \includegraphics[width=8cm, height=8cm]{Exercise11_35_8.eps}
        \end{minipage}
\end{figure}
\section*{VIII. Exercise 11.38 Derive the modified equation of the Lax-Wendroff method for the advection equation.}

The Lax-Wendroff method has the form
\[
	U_j^{n+1}=U_j^{n}-\frac{\mu}{2}\left( U_{j+1}^{n}-U_{j-1}^{n} \right)+\frac{\mu^2}{2}\left( U_{j+1}^{n}-2U_{j}^{n}+U_{j-1}^{n} \right)  
.\] 
For the sake of convenience, we denote $v\left( x, t \right) $ as $v$.
Replace $U_{j}^{n}$ with $v\left( x_j, t_n \right) $ and move the elements on the right side to the left side, we have
\begin{equation*}
	\begin{split}
		0=&\frac{v\left( x, t+k \right)-v\left( x, t \right)  }{k}+\frac{a}{2h}\left[v\left( x+h, t \right)-v\left( x-h, t \right)   \right]
		-\frac{a^2k}{2h^2}\left[ v\left(x+h, t  \right)-2v\left( x, t \right)+v\left( x-h, t \right)    \right]\\
		=&\left[v_t+\frac{k}{2}v_{tt}+\frac{k^2}{6}v_{t t t}+\frac{k^{3}}{24}v_{t t t t} + \mathrm{O}\left( k^{5} \right)\right]
		+\frac{a}{2h}\left[2hv_x+\frac{h^{3}}{3}v_{x x x}+\mathrm{O}\left( h^{5} \right) \right]\\
		 &-\frac{a^2k}{2h^2}\left[h^2v_{x x}+\frac{h^{4}}{12}v_{x x x x}+\mathrm{O}\left( h^{6} \right) \right]\\
		=&v_t+av_x+\frac{k}{2}v_{t t}+\frac{k^2}{6}v_{t t t}+\frac{k^{3}}{24}v_{t t t t}+\frac{ah^2}{6}v_{x x x}-\frac{a^2k}{2}v_{x x}-\frac{a^2h^2k}{24}v_{x x x x}
		+\mathrm{O}\left( k^{4}+h^{4} \right),\\ 
	\end{split}
\end{equation*}
The second line is achieved by Taylor expansion and the third line only sorted the RHS.
Thus we have
\begin{equation}
	\label{eq1}
	v_{t} = -av_{x} + \left( -\frac{k}{2}v_{tt} + \frac{a^2k}{2}v_{x x} \right) + \left(-\frac{k^2}{6}v_{t t t}-\frac{ah^2}{6}v_{x x x}  \right)+\left( \frac{a^2h^2k}{24}v_{x x x x}-\frac{k^{3}}{24}v_{t t t t} \right)
	+\mathrm{O}\left( h^{4} + k^{4} \right) 
\end{equation}
Use this equation recursively to transform all the partial derivatives about t into the one about x, we have
\begin{equation*}
	\begin{split}
		v_{t t} &= -av_{x t} + \left(-\frac{k}{2}v_{t t t} + \frac{a^2k}{2}v_{x x t}  \right)
		+\left( -\frac{k^2}{6}v_{t t t t} - \frac{ah^2}{6}v_{x x x t} \right)+\mathrm{O}\left( h^2k + k^{3} \right)\\
			&=a^2v_{x x} + \left( \frac{a^{3}k}{2}v_{x x x}-\frac{a^{3}k}{2}v_{x x x} \right)
			+\left( \frac{k^2}{4}v_{t t t t}-\frac{a^2k^2}{4}v_{x x t t}-\frac{k^2}{6}v_{t t t t}-\frac{ah^2}{6}v_{x x x t} \right) + \mathrm{O}\left( h^2k + k^{3} \right)\\
			&=a^2v_{x x} + \left( -\frac{a^{4}k^2}{6}v_{x x x x} + \frac{a^2h^2}{6}v_{x x x x} \right)
			+\mathrm{O}\left( h^2k+k^{3} \right).\\ 
	\end{split}
\end{equation*}
Similarly, we have
\begin{equation*}
	\begin{split}
		v_{t t t} &=-a^{3}v_{x x x} + \mathrm{O}\left( k^2 + h^2 \right),\\
		v_{t t t t} &= -\frac{a^{4}}{24}v_{x  x x x} + \mathrm{O}\left( k^2 + h^2 \right).\\ 
	\end{split}
\end{equation*}
Put these equations back to the equation (\ref{eq1}), we have
\begin{equation*}
	\begin{split}
		v_t + av_x =& \left( -\frac{a^2k}{2}v_{x x} + \frac{a^2k}{2}v_{x x} \right)
		+\left( \frac{a^{3}k}{6}v_{x x x} - \frac{ah^2}{6}v_{x x x}\right)\\ 
			    &+ \left(\frac{a^{4}k^{3}}{12}v_{x x x x}
			   -\frac{a^2h^2k}{12}v_{x x x x} + \frac{a^2h^2k}{24}v_{x x x x}  - \frac{a^{4}h^{3}}{24}v_{x x x x}\right)
		+\mathrm{O}\left( h^{4} + k^{4} \right)\\
		=&\frac{a^{3}k^2-ah^2}{6}v_{x x x} + \epsilon_wv_{x x x x} + \mathrm{O}\left( h^{4} + k^{4} \right),
	\end{split}
\end{equation*}
where $\epsilon_w = \mathrm{O}\left( k^2+h^2 \right) $.
Omit the high-order terms and we get the modified equation as
\[
	v_{t} + av_{x}+\frac{ah^2}{6}\left(1-\mu^2  \right)v_{x x x} = 0 
.\]

\section*{IX. Exercise 11.40 Show the modified equation of the leapfrog method.}
For the Lax-Wendroff method, as we have shown in section VIII, we have
 \[
	 v_{t} + av_{x} + \frac{ah^2}{6}\left( 1-\mu^2 \right)v_{x x x} = \epsilon_wv_{x x x x} 
.\] 
In the same way, we replace $U_j^{n}$ with $v\left( x_j, t_n \right) $, we have
\begin{equation*}
	\begin{split}
		0&=\frac{v\left( x, t+k \right)-v\left( x, t-k \right)  }{2k} 
		+ \frac{a v\left( x+h, t \right) - av\left( x - h, t \right) }{2h}\\
		 &=v_t + \frac{k^2}{6}v_{t t t} + \frac{k^{4}}{120}v_{t t t t t} + \mathrm{O}\left( k^{6} \right)
		 +av_{x} + \frac{ah^2}{6}v_{x x x} + \frac{ah^{4}}{120}v_{x x x x x} + \mathrm{O}\left( h^{6} \right).\\ 
	\end{split}
\end{equation*}
Thus we have
\begin{equation}
	\label{eq2}
	\begin{split}
	v_{t} = -av_x + \left( -\frac{k^2}{6}v_{t t t} - \frac{ah^2}{6}v_{x x x} \right)
	+\left( -\frac{k^{4}}{120}v_{t t t t t} - \frac{ah^{4}}{120}v_{x x x x x} \right) + \mathrm{O}\left( h^{6} + k^{6} \right).  
\end{split}
\end{equation}
Like the way we calculate on Lax-Wendroff method, we use equation (\ref{eq2}) recursively to eliminate all the derivatives about t, we have
\begin{equation*}
	\begin{split}
		v_{t t t} &= -a^{3}v_{x x x} + \left( \frac{a^{5}k^2}{6}v_{x x x x x} - \frac{a^{3}h^2}{6}v_{x x x x x} \right)+\mathrm{O}\left( k^{4} + h^{4} \right),\\
		v_{t t t t t} &= -a^{5}v_{x x x x} + \mathrm{O}\left( k^2 + h^2 \right).\\ 
	\end{split}
\end{equation*}
Put these equations back to equation (\ref{eq2}), we have
\begin{equation*}
	\begin{split}
		v_t + av_x =& \frac{a^{3}k^2}{6}v_{x x x} - \frac{ah^2}{6}v_{x x x}\\
			    &+ \left(-\frac{a^{5}k^{4}}{36} + \frac{a^{3}h^2k^2}{36} + \frac{a^{5}k^{4}}{120} - \frac{ah^{4}}{120}  \right)v_{x x x x x} + \mathrm{O}\left( h^{6} + k^{6} \right)\\
	\end{split}
\end{equation*}
Omit the high-order terms and we get the modified equation as
\[
	v_{t} + av_{x} +\frac{ah^2}{6}\left( 1-\mu^2 \right)v_{x x x} = \epsilon_fv_{x x x x x} 
,\] 
where $\epsilon_f=\mathrm{O}\left( k^{4} + h^2k^2 + h^{4} \right) $. Thus if we also omit the $v_{x x x x x}$ term, 
we get (11.37).

\section*{X Exercise 11.41 Show the modified equation of the Beam-Warming method for $a\ge 0$.}
In this exercise, we only discuss the situation when $a\ge 0$.
The Beam-Warming method is
\[
	\frac{U_j^{n+1}-U_j^{n}}{k}=-\frac{a}{2h}\left( 3U_j^{n} - 4U_{j-1}^{n}+U_{j-2}^{n} \right)
	+\frac{a^2k}{2h^2}\left( U_{j}^{n}-2U_{j-1}^{n}+U_{j-2}^{n} \right)  
.\] 
Repalce the $U_j^{n}$ with $v\left( x_j, t_n \right) $ and we have
\begin{equation*}
	\begin{split}
		0 =& \frac{v\left( x, t+k \right) - v\left( x, t \right)  }{k} +
		\frac{a}{2h}\left[ 3v\left( x, t \right) - 4v\left( x - h, t \right)+v\left( x-2h, t \right)    \right]\\
		   &-\frac{a^2k}{2h^2}\left[ v\left( x, t \right)-2v\left( x-h,t \right)+v\left( x-2h, t \right)    \right]\\ 
		=&v_t + \frac{k}{2}v_{t t} + \frac{k^2}{6}v_{t t t} + \mathrm{O}\left( k^{3} \right)
		+\frac{a}{2h}\left(4hv_x - \frac{4h^2}{2}v_{x x} + 
		\frac{4h^{3}}{6}v_{x x x} - 2hv_x + \frac{4h^2}{2}v_{x x}-\frac{8h^{3}}{6}v_{x x x}+\mathrm{O}\left( h^{4} \right)   \right)\\ 
	\end{split}
\end{equation*}
Sorted is and we have
\begin{equation}
	\label{eq3}
	v_t = -av_x+\left( -\frac{k}{2}v_{t t}+\frac{a^2k}{2}v_{x x} \right)+
	\left( -\frac{k^2}{6}v_{t t t} + \frac{ah^2}{3}v_{x x x}-\frac{a^2hk}{2}v_{x x x} \right)
	+\mathrm{O}\left( h^{3}+k^{3}+h^2k \right) 
\end{equation}
Use equation (\ref{eq3}) recursively to eliminate all the derivatives about t, we have
\begin{equation*}
	\begin{split}
		v_{t t} =& a^2v_{x x}+\left( \frac{ak}{2}v_{t t x} - \frac{a^{3}k}{2}v_{x x x}-\frac{k}{2}v_{t t}+\frac{a^2k}{2}v_{x x} \right)
		+\mathrm{O}\left( k^2+hk+h^2 \right)\\
		=&a^2v_{x x}+\left( \frac{a^{3}k}{2}v_{x x x}-\frac{a^{3}k}{2}v_{x x x}-\frac{a^2k}{2}v_{x x}+\frac{a^2k}{2}v_{x x} \right) + \mathrm{O}\left( k^2+hk+h^2 \right)\\
		=&a^2v_{x x} + \mathrm{O}\left( h^2+hk+k^2 \right),\\
		v_{t t t} =& -a^{3}v_{x x x} + \mathrm{O}\left( k \right).\\ 
	\end{split}
\end{equation*}
Putting these equation back to equation (\ref{eq3}) gives
\begin{equation*}
	\begin{split}
		v_t+av_x =& \left( -\frac{a^2k}{2}v_{x x}+\frac{a^2k}{2}v_{x x} \right)
		+\left( \frac{a^{3}k^2}{6}v_{x x x} + \frac{ah^2}{3}v_{x x x}-\frac{a^2hk}{2}v_{x x x} \right)
		+\mathrm{O}\left( h^{3}+k^{3}+h^2k \right)\\
	\end{split}
\end{equation*}
Neglect the high-order terms and we have the modified equation as
\[
	v_t + av_x + \frac{ah^2}{6}\left( -2+3\mu-\mu^2 \right)v_{x x x} = 0 
.\] 

Thus we have
\begin{equation*}
	\begin{split}
		C_p\left( \xi \right) &=a+\frac{ah^2}{6}\left( \mu-1 \right)\left(\mu-2  \right) \xi^2,\\
		C_g\left( \xi \right)&=a+\frac{ah^2}{2}\left( \mu-1 \right)\left( \mu-2 \right)\xi^2.\\   
	\end{split}
\end{equation*}
In the Example 11.34, $\mu=0.8$ and $a=1$. 
We can easily find that both $C_p$ and $C_g$ have a magnitude bigger than $|a|$. 
Because $\frac{h^2}{6}\left( 0.8-1 \right)\left( 0.8-2 \right)\xi^2> 0  $
when $\xi \neq  0$.
Hence the numerical oscillations step forward the true wave crest; this answers Question (e) of Example 11.34.

\section*{XI. Exercise 11.42 What if $\mu=1$?}
When $\mu=1$, obviously we have  $a> 0$.
In the case when $\mu=1$, we can write out the equation about Lax-Wendroff method, which is
\begin{equation*}
	\begin{split}
		U_{j}^{n+1}=&U_j^{n}-\frac{1}{2}\left( U_{j+1}^{n}-U_{j-1}^{n} \right)+
		\frac{1}{2}\left( U_{j+1}^{n}-2U_{j}^{n}+U_{j-1}^{n} \right)\\ 
		=&U_{j-1}^{n}
	\end{split}
\end{equation*}
This means Lax-Wendroff method only moves the solution $U^{n}$ along the deriction of velocity $a$ over a distance of $h=ak$. 
In another words, if the $U^{n}$ is the exact solution, so is $U^{n+1}$.
In Example 11.34, we take the initial condition $U^{0}$, which is the exact solution when $t=0$, 
thus the final solution when  $T=17$ is also the exact solution when we use Lax-Wendroff method.

When $\mu=1$, the leapfrog method is
\begin{equation*}
	\begin{split}
		U_j^{n+1}-U_j^{n-1}&=-\mu\left( U_{j+1}^{n}-U_{j-1}^{n}\right)\\
				   &=-U_{j+1}^{n}+U_{j-1}^{n}.\\
	\end{split}
\end{equation*}
If $U^{n}$ and $U^{n-1}$ are all exact solution, we easily have $U_j^{n-1}=U_{j+1}^{n}$, where $j+1=1$ if $j=m+1$.
Thus we have
\[
	U_j^{n+1}=U_{j-1}^{n}
.\] 
Which means this method also moves the solution $U^{n}$ along the deriction of velocity $a$ over a distance of $h=ak$.
So $U^{n+1}$ is also a exact solution.
Thus if we use exact solution to set initial condition $U^{0}$ and $U^{1}$, 
we have the final solution for $T=17$ is also the exact solution. 

Thus we answered Question (f) why the numerical result is so good when $\mu = 1$ for Lax-Wendroff method and leapfrog method.

\section*{XII. Exercise 11.43 Apply the von Neumann analysis to the upwind method to derive its amplification factor.}
We set
\[
	U_j^{n}=\left[g\left( \xi \right) \right]^{n}e^{ix_j\epsilon}
.\] 
And we expect that
 \[
	 U_j^{n+1}=g\left( \xi \right)U_j^{n} 
.\] 
Inserting these expressions into the equation about upwind method for $a\ge 0$, we have
\begin{equation*}
	\begin{split}
		g\left( \xi \right)U_j^{n+1}=U_j^{n+1}
		&=U_j^{n}-\mu\left( U_j^{n}-U_{j-1}^{n} \right)\\
		&=U_j^{n}-\mu\left( U_j^{n}-e^{-ih\xi}U_j^{n} \right)\\
		&=\left( 1-\mu+\mu e^{-ih\xi} \right)U_j^{n}.\\ 
	\end{split}
\end{equation*}
The last line is achieved by Euler's formula.
Thus the amplification factor of upwind method for $a\ge 0$ is
\[
	g\left( \xi \right)=\left( 1-\mu \right)+\mu e^{-ih\xi}  
.\]

In order to let the method be stable, we need to guarantee that 
\begin{equation*}
	\begin{split}
		\quad\mid g\left( \xi \right) \mid\le 1
		\Rightarrow \left( 1-\mu+\mu \cos\left( h\xi \right)  \right)^2+\mu^2\sin^2\left( h\xi \right)\le 1\\
	\end{split}
\end{equation*}
Let $x=\cos\left( h\xi \right) $, which will range form $-1$ to $1$, and we want to have
\begin{equation*}
 	\begin{split}
		\mid g\left( \xi \right) \mid =f\left( x, \mu \right)&=\left( 1-\mu+\mu x \right)^2+\mu^2\left( 1-x^2 \right)\\
		&=\left(2\mu-2\mu^2  \right)x+\mu^2+\left( 1-\mu \right)^2  \\
		&\le 1 \quad \forall x\in\left[-1, 1\right].\\						     
 	\end{split}
\end{equation*}
For $a\ge 0$, it is obvious that $\mu\ge 0$.

When $0\le \mu\le 1$, we have  $2\mu-2\mu^2\ge 0$, which means
\begin{equation*}
	\begin{split}
		\max_{x\in\left[-1, 1\right]}f\left( x, \mu \right)&=f\left( 1, \mu \right)
		=2\mu-2\mu^2+\mu^2+\left( 1-\mu \right)^2
		=1
	\end{split}
\end{equation*}

When $\mu>1$, we have  $2\mu-2\mu^2<0$, which means
\begin{equation*}
	\begin{split}
		\max_{x\in\left[-1, 1\right]}f\left( x, \mu \right)&=f\left( 1, \mu \right)
		=-2\mu+2\mu^2+\mu^2+\left( 1-\mu \right)^2
		=\left( 2\mu-1 \right)^2
		> 1
	\end{split}
\end{equation*}

Thus when $0\le \mu\le 1$, the method is stable in the case of $a\ge0$.

\section*{XIII. Exercise 11.44 Apply the von Neumann analysis to the Lax-Fredrichs method to derive its amplification factor.}
Like section XII, we have
\begin{equation*}
	\begin{split}
		g\left( \xi \right)=U_j^{n+1}&=\frac{1}{2}\left( U_{j+1}^{n}+U_{j-1}^{n} \right)
		-\frac{\mu}{2}\left( U_{j+1}^{n}-U_{j-1}^{n} \right)\\
		&=\frac{1}{2}\left( e^{ih\xi}+e^{-ih\xi} \right)U_{j}^{n}-\frac{\mu}{2}\left( e^{ih\xi}-e^{-ih\xi} \right)U_{j}^{n}\\
		&=\left[ \cos\left( h\xi \right)-i\mu\sin\left( h\xi \right)   \right]U_{j}^{n} 
	\end{split}
\end{equation*}
The last line is achieved by Euler's formula.
Thus its amplification factor is
\[
	g\left( \xi \right)=\cos\left( \xi h \right)-\mu i\sin\left( h\xi \right)   
.\]
Let $x=\cos\left( h\xi \right) $, which will range from $-1$ to  $1$.
In order to guarantee that the method is stable we need to let
\begin{equation*}
	\begin{split}
		\mid g\left( \xi \right) \mid =f\left( x, \mu \right)
		&=x^2+\mu^2\left( 1-x^2 \right)\\
		&=\left( 1-\mu^2 \right)x^2+\mu^2\\
		&\le 1\quad \forall x\in\left[-1, 1\right].\\
	\end{split}
\end{equation*}
It is obvious that $f\left( x, \mu \right)\ge 0 $ and $\mu^2\ge 0$.

When $\mu^2\le 1$. We have
\begin{equation*}
	\begin{split}
		\max_{x\in\left[-1, 1\right]}&=f\left( 1, \mu \right)
					     =1-\mu^2+\mu^2
					     =1
	\end{split}
\end{equation*}

When $\mu^2\ge 1$. We have
\begin{equation*}
	\begin{split}
		\max_{x\in\left[-1, 1\right]}&=f\left( 0, \mu \right)
					     =\mu^2
					     >1.
	\end{split}
\end{equation*}

Thus when $-1\le \mu\le 1$, the method is stable.

\section*{XIV. Exercise 11.45. Apply the von Neumann analysis to the Lax-Wendroff method to derive its amplification factor.}
Like section XII, we have
\begin{equation*}
	\begin{split}
		g\left( \xi \right)=U_{j}^{n+1}
		&=U_j^{n}-\frac{\mu}{2}\left( U_{j+1}^{n}-U_{j-1}^{n}\right)
		+\frac{\mu^2}{2}\left( U_{j+1}^{n}-2U_{j}^{n}+U_{j-1}^{n} \right)\\
		&=U_j^{n}-\frac{\mu}{2}\left( e^{ih\xi}-e^{-ih\xi} \right)U_{j}^{n}+
		\frac{\mu^2}{2}\left( e^{ih\xi}-2+e^{-ih\xi} \right)U_{j}^{n}\\
		&=\left[1-i\mu \sin\left( h\xi \right)+\mu^2\left( \cos\left( h\xi \right)-2  \right)  \right]U_{j}^{n}\\
		&=\left[1-2\mu^2\sin^2\left( \frac{h\xi}{2} \right)-i\mu\sin\left( h\xi \right)  \right]U_{j}^{n}.\\
	\end{split}
\end{equation*}
The third line is achieved by Euler's method and the last one is achieved by double angle formula.
Let $x=\sin\left( \frac{h\xi}{2} \right) $, which will range from $-1$ to  $1$.
In order to guarantee that the method is stable, we need to let
\begin{equation*}
	\begin{split}
		\mid g\left( \xi \right) \mid =f\left( x, \mu \right)
		&=\left( 1-2\mu^2x^2 \right)^2+4\mu^2\left( 1-x^2 \right)x^2\\
		&=\left( 4\mu^{4}-4\mu^{2} \right)x^{4}+1 \\
		&\le 1\quad x\in\left[-1, 1\right].\\
	\end{split}
\end{equation*}

When $\mu^2\le 1$, thus $4\mu^{4}-4\mu^2\le 0$. We have
\[
	\max_{x\in\left[-1, 1\right]} \mid g\left( \xi \right) \mid
	=f\left( 0, \mu \right)=1   
.\] 

When $\mu^2>1$, thus $4\mu^{4}-4\mu^2>0$. We have
\[
	\max_{x\in\left[-1, 1\right]} \mid g\left( \xi \right)  \mid 
	=f(\left( 1, \mu \right)=\left( 2\mu^2-1 \right)^2>1  
.\] 

Thus when $-1\le \mu\le 1$, the method is stable.
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